Don't use vapply() here - use lapply() instead and then leave cut's output
alone.
vapply() will combine its outputs to create a character matrix and
data.frame will pull apart the character matrix into its columns. Skipping
the matrix intermediary solves
lots of issues.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Jan 11, 2016 at 11:24 AM, Guelman, Leo <leo.guel...@rbc.com> wrote:
> Dear list,
>
> What is a better way relative to the one below to keep the order of factor
> levels created from cut()? Notice, I'm simply pasting letters to levels
> before converting to character so to keep the desired order of levels. This
> is not very elegant... I'm converting to character so I can call the helper
> fun with vapply() from the main fun.
>
> Removing this line of code " levels(xc) <- paste(letters[1:nlevels(xc)],
> levels(xc), sep=":")" would result in factor levels that are not ordered
> according to x1.
>
> set.seed(1)
> df <- data.frame(x1 = rnorm(1000), x2 = rnorm(1000))
>
> main_fun <- function(data) {
> data.frame(vapply(data, helper_fun, character(nrow(df))))
> }
>
> helper_fun <- function(x) {
> xc <- cut(x, breaks = unique(quantile(x, seq(0, 1, 1/10), na.rm =
> TRUE)),
> include.lowest = TRUE)
> levels(xc) <- paste(letters[1:nlevels(xc)], levels(xc), sep=":")
> as.character(xc)
>
> }
>
>
> res <- main_fun(df)
> levels(res$x1)
> levels(res$x1)
> [1] "a:[-3.01,-1.34]" "b:(-1.34,-0.882]" "c:(-0.882,-0.511]"
> "d:(-0.511,-0.296]" "e:(-0.296,-0.0353]"
> [6] "f:(-0.0353,0.245]" "g:(0.245,0.536]" "h:(0.536,0.854]"
> "i:(0.854,1.32]" "j:(1.32,3.81]"
> >
>
> Thanks
> Leo.
>
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